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S.chand Class 8 Maths Solution Chapter 11 Time And Work Exercise 11

Exercise 11

Question 1

1. 6 men can complete the electric fitting in a building in 7 days. How many days will it take 21 men to do the job?

2. Shashi weaves 25 baskets in 35 days. In how many days will she weave 110 baskets ?

3. Two men can do a piece of work in 3 days and 6 days respectively. If they work together, in how many days will they finish the work?

4. One man can do a piece of work in 3 hours ; another can do the same piece of work in 2 hours. How long will they take if they work together?

5. $A$ and $B$ together can do a piece of work in 10 days, but $A$ alone can do it in 15 days. How many days would $B$ alone take to do the same work ?

6. Two men can do a piece of work in 6 hours and 4 hours respectively. After the first has worked for 2 hours, he is joined by the other. By when should the work be completed ?

7. $A, B$ and $C$ can cultivate a field in 10,12 , and 15 days respectively. If they work together, in how many days will they finish the work and what fraction of the work will each of them do ?

8. $A, B$ and $C$ can do a piece of work in 10 days. $A$ alone can do it in 40 days, and $B$ alone can do it in 30 days. In how many days will $C$ alone do the same work?

9. $A$ and $B$ can do a given piece of work in 8 days; $B$ and $C$ can do the same work in 12 days and $A, B, C$ complete it in 6 days. In how many days can $A$ and $C$ finish it?

10. A cistern can be filled by one tap in 4 hours and by another in 3 hours. How long will it take to fill if both taps are opened together?

11. A cistern can be filled by a tap in 4 hours and emptied by an outlet pipe in 6 hours. How long will it take to fill the cistern if both taps are opened together?

12. One tap fills a bath in $12 \mathrm{~min}$. and another tap fills it in $15 \mathrm{~min}$. The waste-pipe can empty the bath in $10 \mathrm{~min}$. In what time will the bath be filled if both taps are turned on and if the waste-pipe has been left open accidentally ?

13. A pipe can fill a cistern in 6 hours. Due to a leak in the bottom it is filled in 7 hours. When the cistern is full, in how much time will it be emptied by the leak ?

14. A cistern has two inlets $A$ and $B$ which can fill it in 15 hours and 20 hours respectively. An outlet car empty the full cistern in 12 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely?

Multiple Choice Questions (MCQs)

15. $A$ and $B$ together take 5 days to do a work, $B$ and $C$ take 7 days to do the same and $A$ and $C$ take 4 days to do it. Who among these will take the least time, if put to do it alone?

(a) $A$

(b) $B$

(d) data is not adequate

16. A tap can empty a tank in one hour. A second tap can empty it in 30 minutes. If both the taps operate simultaneously, how much time is needed to empty the tank?

(a) 20 min.

(b) 30 min.

(d) 45 min.

High Order Thinking Skills (HOTS)

17. $A$ and $B$ can finish a work in 30 days. They worked for it for 20 days and then $B$ left. The remaining work was done by $A$ alone in 20 more days. $A$ alone can finish the work in

(a) 54 days

(b) 60 days

(d) 50 days

18. Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. The tank is half full. All the three pipes are in operation simultaneously. After how much time will the tank will be full.

S.chand mathematics solution class 8 chapter 11 time and work

EXERCISE 11

Q1 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper

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Question 1

A can do $\dfrac{3}{4}$ of the work in 12 days . In how many days can he complete $\dfrac{1}{8}th$ of the work ?

Sol :

⇒A can do $\dfrac{3}{4}$ of work in 12 days

⇒ A can do 1 work in $12\times \dfrac{4}{3}$ days And [Multiplying both side by 4/3]

⇒ A can do $\dfrac{1}{8}$ work in $12\times \dfrac{4}{3}\times \dfrac{1}{8}$ days [Multiplying both side by 1/8]

$=4\times \dfrac{1}{2}$ days

= 2 days

Q2 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper

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Question 2

Tanu can do a piece of work in 24 days and Manisha can do it in 30 days . How long will they take to do the work working together ?

Sol :

⇒Tanu can do a piece of work in 24 days

⇒So Tanu's one day work $=\dfrac{1}{24}$

⇒Manisha can do the same work in 30 days

⇒So Manisha's one day work $=\dfrac{1}{30}$

⇒We know that

⇒Tanu and Manisha's one day work = Tanu's one day work + Manisha's one day work

$=\dfrac{1}{24}+\dfrac{1}{30}$ [Taking LCM]

$=\dfrac {5 + 4}{120}$

$= \dfrac{9}{120}$

$= \dfrac{3}{40}$

⇒Tanu and Manisha's one day work $= \dfrac{3}{40}$

⇒Total time required for completing work $\dfrac{1}{\text{one day work}}=\dfrac{1}{\dfrac{3}{40}}$

⇒So Tanu and Manisha can complete the same work in $= \dfrac{40}{3}=13\dfrac{1}{3}$ days.

Q3 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper

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Question 3

A and B can polish the floors of a building in 18 days . A alone can do $\dfrac{1}{6}$ of this job in 4 days . in how many days can B alone polish the floors ?

Sol :

⇒A alone complete whole work in $=\dfrac{4}{\dfrac{1}{6}}=4\times\dfrac{6}{1}$ = 24 days

⇒A's one day work $=\dfrac{1}{24}$

⇒(A and B)'s one day work $=\dfrac{1}{18}$

⇒Also , (A+B)'s one day work = A's one day work + B's one day work

⇒$\dfrac{1}{18}=\dfrac{1}{24}+\dfrac{1}{x}$

⇒$\dfrac{4-3}{72}=\dfrac{1}{x}$

⇒$\dfrac{1}{72}=\dfrac{1}{x}$

⇒x = 72

72 days

Q4 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper

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Question 4

A and B can weave a certain number of baskets in 12 days and 15 days respectively . They work together for 5 days and then B leaves . In how many days will A finish the rest of the work ?

Sol :

⇒A's one day work $=\dfrac{1}{12}$

⇒B's one day work $=\dfrac{1}{15}$

⇒(A + B)'s one day work = A's one day work + B's one day work

$=\dfrac{1}{12}+\dfrac{1}{15}$

$=\dfrac{5+4}{60}$

$=\dfrac{9}{60}$

$=\dfrac{3}{20}$

⇒(A + B)'s one day work $=\dfrac{3}{20}$

⇒They work together for 5 days $=5\times \dfrac{3}{20}= \dfrac{3}{4}$

⇒Remaining work $=1-\dfrac{3}{4}$

$=\dfrac{4-3}{4}=\dfrac{1}{4}$

⇒A can complete all work in 12 days . So , remaining work can be done in $=\dfrac{1}{4}\times 12$ days

= 3 days

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Question 5

A , B and C working together take 8 min . to address a pile of envelopes . A and B together would take 10 min ; A and C together would take 15 min . How long would each take working alone ?

Sol :

⇒(A + B + C)'s one unit work $=\dfrac{1}{8}$

⇒(A + B)'s one unit work $=\dfrac{1}{10}$

⇒(A + C)'s one unit work $=\dfrac{1}{15}$

⇒C's one unit work = (A + B + C)'s one unit work - (A + B)'s one unit work

$=\dfrac{1}{8}-\dfrac{1}{10}$

$=\dfrac{5-4}{40}=\dfrac{1}{40}$

∴C alone take 40 min

⇒B's one unit work = (A + B + C)'s one unit work - (A + C)'s one unit work

$=\dfrac{1}{8}-\dfrac{1}{15}$

$=\dfrac{15-8}{120}=\dfrac{7}{120}$

∴B alone take $\dfrac{120}{7}$ min

⇒A's one unit work = (A + B + C)'s one unit work - B's one unit work - C's one unit work

$=\dfrac{1}{8}-\dfrac{7}{120}-\dfrac{1}{40}$

$=\dfrac{15-7-3}{120}=\dfrac{5}{120}=\dfrac{1}{24}$

∴A alone take 24 min

A→24 min ; B→$\dfrac{120}{7}$ min ; C→40 min

Q6 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper

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Question 6

A and B can do a piece of work in 6 days and 4 days respectively . A started the work ; worked at it for 2 days and then was joined by B . Find the total time taken to complete the work .

Sol :

⇒A's one day work $=\dfrac{1}{6}$

⇒B's one day work $=\dfrac{1}{4}$

⇒A's 2 days work $=2\times \dfrac{1}{6}=\dfrac{1}{3}$

⇒Remaining work $=1-\dfrac{1}{3}$

$=\dfrac{3-1}{3}=\dfrac{2}{3}$..(i)

⇒(A+B)'s one day work = A's one day work + B's one day work

$=\dfrac{1}{6}+\dfrac{1}{4}$

$=\dfrac{2+3}{12}=\dfrac{5}{12}$

⇒Total time required to complete work together$=\dfrac{1}{\text{one day work}}$ $=\dfrac{1}{\dfrac{5}{12}}=\dfrac{12}{5}$

⇒Time to complete the remaining work together $=\dfrac{2}{3}\times \dfrac{12}{5}$ $=\dfrac{8}{5}$..(ii)

⇒Total time required to complete work = A's alone working time + (A+B)'s working together time

$=2+\dfrac{8}{5}$

$=\dfrac{10+8}{5}$

$=\dfrac{18}{5}=3\dfrac{3}{5}$

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Question 7

780 is paid for a task which A can do in 6 days , B in 8 days and C in 4 days . If all work together , how much money should each receive ?

Sol :

⇒A's one day work $=\dfrac{1}{6}$

⇒B's one day work $=\dfrac{1}{8}$

⇒C's one day work $=\dfrac{1}{4}$

∴Ratio of shares of money of A , B and C = Ratio of their one day work

$=\dfrac{1}{6}:\dfrac{1}{8}:\dfrac{1}{4}$ $=\dfrac{1}{6}\times 24:\dfrac{1}{8}\times 24 :\dfrac{1}{4}\times 24$ [L.C.M of 6 , 8 and 24]

= 4 : 3 : 6

∴Sum of the ratios = 4 + 3 + 6 = 13

⇒A's share $=\dfrac{4}{13}\times 780$ = 240

⇒B's share $=\dfrac{3}{13}\times 780$ = 180

⇒C's share $=\dfrac{6}{13}\times 780$ = 360

A→240 , B→180 , C→360

ALTERNATE METHOD

⇒A's one day work $=\dfrac{1}{6}$

⇒B's one day work $=\dfrac{1}{8}$

⇒C's one day work $=\dfrac{1}{4}$

Total work done$=\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{4}$

$=\dfrac{4+3+6}{24}=\dfrac{13}{24}$

∴Shares of money of A , B and C = Ratio of their one day work / Total work × Total money

⇒A's share $\dfrac{\frac{1}{6}}{\frac{13}{24}}\times 780$

$=\dfrac{4}{13}\times 780$ = 240

⇒B's share $\dfrac{\frac{1}{8}}{\frac{13}{24}}\times 780$

$=\dfrac{3}{13}\times 780$ = 180

⇒C's share $\dfrac{\frac{1}{4}}{\frac{13}{24}}\times 780$

$=\dfrac{6}{13}\times 780$ = 360

A→240 , B→180 , C→360

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Question 8

X works thrice as fast as Y . If Y alone can complete a job in 18 days , find how ,many days will X and Y together take to complete the job .

Sol :

⇒Y's one day work $=\dfrac{1}{18}$

A.T.Q

⇒3X=Y

⇒$X=\dfrac{Y}{3}$

⇒$X=\dfrac{18}{3}$

⇒X=6

⇒X's one day work $=\dfrac{1}{6}$

⇒(X+Y)'s one day work = X's one day work + Y's one day work

$=\dfrac{1}{6}+\dfrac{1}{18}$

$=\dfrac{3+1}{18}=\dfrac{4}{18}=\dfrac{2}{9}$

⇒Total time required by (X+Y) to complete work $=\dfrac{1}{\text{one day work}}$ $=\dfrac{1}{\dfrac{2}{9}}$ $=\dfrac{9}{2}$ or $=4\dfrac{1}{2}$ days

Q9 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper

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Question 9

A can do a job in 20 days , B in 30 days and C in 60 days . If A is helped by B and C on every third day , then in how many days will the job be finished ?

Sol :

⇒A's one day work $=\dfrac{1}{20}$..(i)

⇒(A+B+C)'s one day work = A's one day work + B's one day work + C's one day work

$=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{60}$

$=\dfrac{3+2+1}{60}$

$=\dfrac{6}{60}=\dfrac{1}{10}$..(i)

⇒According to question, A work starting two days alone and on every third day all together work .

Total time to complete work = 1st day + 2nd day + 3rd day

1 and 2 day equal to (i) and 3 day equal to (ii)

$=\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{10}$

$=\dfrac{1+1+2}{20}=\dfrac{4}{20}=\dfrac{1}{5}$

Which means in 3 days 1/5 work is completed

$\text{3 days}=\dfrac{1}{5}\text{ work}$

$3\times 5 = 1 \text{ work}$

= 15 days to complete whole work

Q10 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper

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Question 10

Sonia can copy 50 pages in 10 hours . Priya and Sonia can copy 300 pages in 40 hours . In how much time can Priya copy 30 pages ?

Sol :

⇒Sonia in 10 hours can copy 50 pages

∴In 1 hour sonia can copy $\dfrac{50}{10}=\dfrac{5}{1}$ pages

⇒Priya and Sonia together in 40 hours can copy 300 pages

∴Priya and Sonia together in 1 hours can copy $\dfrac{300}{40}=\dfrac{30}{4}$ pages

⇒Priya's 1 hour work = Priya and Sonia together in 1 hours work - Sonia's 1 hour work

$=\dfrac{30}{4}-\dfrac{5}{1}$

$=\dfrac{30-20}{4}=\dfrac{10}{4}=\dfrac{5}{2}$

⇒Number of hours required to complete the work $=\dfrac{\text{Work to be done}}{\text{one day work}}$

$=\dfrac{30}{\dfrac{5}{2}}$

$=5\times\dfrac{2}{5}$

= 12 hours

Q11 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper

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Question 11

A tap can fill a cistern in 15 minutes and another can empty it in 18 minutes . Find in how many minutes the tank will be filled up ?

Sol :

⇒Tank filled by tap in one hour $=\dfrac{1}{15}$ part

⇒Tank empty by tap in one hour $=\dfrac{1}{18}$ part

⇒Tank filled in one hour when both taps are open $=\dfrac{1}{15}-\dfrac{1}{18}$

$=\dfrac{6-5}{90}=\dfrac{1}{90}$

= 90 mins the tank will be filled up

Q12 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper

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Question 12

Two taps can fill a cistern in 15 hours and 20 hours respectively . A third tap can empty it in 30 hours . How long will they take to fill the cistern if all the taps are opened ?

Sol :

⇒Tank filled by first tap in one hour $=\dfrac{1}{15}$ part

⇒Tank filled by second tap in one hour $=\dfrac{1}{20}$ part

⇒Tank empty by third tap in one hour $=\dfrac{1}{30}$ part

⇒Tank filled in one hour when all three taps are open $=\dfrac{1}{15}+\dfrac{1}{20}-\dfrac{1}{30}$

$=\dfrac{4+3-2}{60}=\dfrac{5}{60}=\dfrac{1}{12}$

= 12 hours take the tank to fill up

Q13 | Ex-11 | Cl-8 | Schand Composite Mathematics | Time and Work | Myhelper

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Question 13

Two taps running together can fill a bath in 4 minutes , which is filled by one of the taps by itself in 7 minutes . How long would it take if the other pipe is running by itself ?

Sol :

⇒Bathtub filled by (1st + 2nd) taps in one hour $=\dfrac{1}{4}$ part

⇒Bathtub filled by first tap in one hour $=\dfrac{1}{7}$ part

⇒Bathtub filled by (1st + 2nd) taps in one hour = time taken by 1st tap in one hour + time taken by 2nd tap in one hour

⇒$\dfrac{1}{4}=\dfrac{1}{7}+x$

⇒$\dfrac{1}{4}-\dfrac{1}{7}=x$

⇒$\dfrac{7-4}{28}=x$

⇒$x=\dfrac{3}{28}$

Time taken by Second tap $=\dfrac{28}{3}=9\dfrac{1}{3}$ minutes

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Question 14

A cistern can be filled by 3 taps A, B and C when turned on separately in 12 min , 10 min and 15 min respectively . If all are turned on together for $2\dfrac{2}{3}$ minutes and if B and C are then turned off , how much time will A alone take to fill the cistern ?

Sol :

⇒Time taken when all 3 taps are open for a unit time = Tap A open for unit time + Tap B open for unit time + Tap C open for unit time

$=\dfrac{1}{12}+\dfrac{1}{10}+\dfrac{1}{15}$

$=\dfrac{5+6+4}{60}=\dfrac{15}{60}=\dfrac{1}{4}$

A.T.Q

⇒All are turned for $2\dfrac{2}{3}$ minutes $=\dfrac{1}{4}\times 2\dfrac{2}{3}$ $=\dfrac{1}{4}\times \dfrac{8}{3}=\dfrac{2}{3}$

⇒Remaining cistern to be filled $=1-\dfrac{2}{3}$ $=\dfrac{3-2}{3}=\dfrac{1}{3}$

⇒Tap A alone fill the remaining cistern [1/3 of 12min] $=\dfrac{1}{3}\times 12$

= 4 minutes

RS Aggarwal solution class 8 chapter 13 Time and Work Test Paper 13

Test Paper 13

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Q1 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 1:

A can do a piece of work in 10 days while B alone can do it in 15 days. In how many days can both finish the same work?

Answer 1:

$A can do a piece of work in 10 days . A' s 1 day work = \frac{1}{10} B can do a piece of work in 15 days . B' s 1 day work = \frac{1}{15} (A + B)' s 1 day work = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} A and B working together can complete the work in 6 days .$

Q2 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 2:

A and B can do a piece of work in 15 days; B and C in 12 days; C and A in 20 days. How many days will be taken by A, B and C working together to finish the work?

Answer 2:

$(A + B) can do a work in 15 days . ∴ (A + B)' s 1 day work = \frac{1}{15} (B + C) can do a work in 12 days . ∴ (B + C)' s 1 day work = \frac{1}{12} (C + A) can do a work in 20 days . ∴ (C + A)' s 1 day work = \frac{1}{20} 2 (A + B + C)' s 1 day work = \frac{1}{15} + \frac{1}{12} + \frac{1}{20} = \frac{4 + 5 + 3}{60} = \frac{12}{60} = \frac{1}{5} (A + B + C)' s 1 day work = \frac{1}{10} A, B and C working together require 10 days to complete the work .$

Q3 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 3:

Tap A can fill a cistern in 8 hours and tap B can empty it in 12 hours. How long will it take to fill the cistern if both of them are opened together?

Answer 3:

$Tap A can fill a cistern in 8 hours . Part of cistern filled by Tap A in 1 hour = \frac{1}{8} Tap B empties the cistern in 12 hours . Part of cistern emptied by Tap B in 1 hour = - \frac{1}{12} (negative sign shows that tap B drains the tank) Part of cistern filled in one hour when both taps are opened together = \frac{1}{8} - \frac{1}{12} = \frac{3 - 2}{24} = \frac{1}{24} Therefore, it will take 24 hours to fill the cistern .$

Q4 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 4:

2 men or 3 women can do a piece of work in 16 days. In how many days can 4 men and 6 women do the same work?

Answer 4:

$Work of 2 men = Work of 3 women \Rightarrow Work of 1 man = \frac{3}{2} women Three women can do a piece of work in 16 days . As 4 men and 6 women = (4 \times \frac{3}{2}) women + 6 women = 6 women + 6 women = 12 women Also, 3 women can do the work in 16 days . So, work done by 3 women in one day = \frac{1}{16} ∴ Work done by 1 woman in one day = \frac{1}{48} \Rightarrow Work done by 12 women in one day = \frac{1}{4} Thus, 4 men and 6 women will complete the work in 4 days .$

Q5 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 5:

A pipe can fill a cistern in 9 hours. Due to a leak in its bottom, the cistern fills up in 10 hours. If the cistern is full, in how much time will it be emptied by the leak?

Answer 5:

$Time taken by the pipe to fill the cistern = 9 hours Part of the cistern filled in one hour = \frac{1}{9} Suppose the leak empties the full cistern in x hours . Part of the cistern emptied in one hour = - \frac{1}{x} (negative sign implies a leak) Time taken by the cistern to fill completely due to the leak = 10 hours Part of the cistern filled in one hour due to the leak = \frac{1}{10} ∴ \frac{1}{10} = \frac{1}{9} - \frac{1}{x} \Rightarrow \frac{1}{x} = \frac{1}{9} - \frac{1}{10} = \frac{1}{90} x = 90 hours Therefore, the leak will empty a full cistern in 90 hours .$

Q6 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 6:

Mark (✓) against the correct answer:
The rates of working of two tapes A and B are in the ratio 2 : 3. The ratio of the time taken by A and B respectively to fill a given cistern is
(a) 2 : 3
(b) 3 : 2
(c) 4 : 9
(d) 9 : 4

Answer 6:

(b) 3:2

$Rates at which taps A and B work = 2 : 3 The ratio of time taken by taps A and B to fill the cistern = \frac{1}{Rates at which taps A and B work} = \frac{1}{\frac{2}{3}} = \frac{3}{2} = 3 : 2$

Q7 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 7:

Mark (✓) against the correct answer:
A can finish a piece of work in 12 hours while B can finish it in 15 hours. How long will both take to finish it, working together?
(a) 9 hours
(b) $6 \frac{2}{3}$ hours
(c) $6 \frac{3}{4}$ hours
(d) $8 \frac{1}{3}$ hours

Answer 7:

(b) $6 \frac{2}{3} h o u r s$

$A' s 1 hour work = \frac{1}{12} B' s 1 hour work = \frac{1}{15} (A + B)' s 1 hour work = \frac{1}{12} + \frac{1}{15} = \frac{9}{60} = \frac{3}{20} Time taken by A and B to complete the work together = \frac{20}{3} = 6 \frac{2}{3} hours$

Q7 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 8:

Mark (✓) against the correct answer:
A can do a piece of work in 14 days and B is 40% more efficient than A. In how many days can B finish it?
(a) 10 days
(b) $7 \frac{1}{2}$ days
(c) $5 \frac{1}{4}$ days
(d) $5 \frac{3}{5}$ days

Answer 8:

(a) 10 days
$A' s 1 day work = \frac{1}{14} B is 40 % more efficient than A . ∴ B' s 1 day work = \frac{140}{100} \times \frac{1}{14} = \frac{1}{10} B takes 10 days to complete the work .$

Q9 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 9:

Mark (✓) against the correct answer:
A pump can fill a cistern in 2 hours. Due to a leak in the tank it takes $2 \frac{1}{3}$ hours to fill it. The leak can empty the full tank in
(a) 7 hours
(b) 14 hours
(c) 8 hours
(d) 3 hours

Answer 9:

(b) 14 hours

$A pump fills a tank in 2 hours . Part of tank filled by the pump in one hour = \frac{1}{2} Let x hours be the time required for water to leak from the cistern . Part of tank drained by the leak in one hour = - \frac{1}{x} Time required to fill the leaking cistern = 2 \frac{1}{3} = \frac{7}{3} hours Part of tank stored with water in one hour = \frac{3}{7} \Rightarrow \frac{3}{7} = \frac{1}{2} - \frac{1}{x} \frac{1}{x} = \frac{1}{2} - \frac{3}{7} = \frac{7 - 6}{14} = \frac{1}{14} x = 14 hours$

Q10 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 10:

Mark (✓) against the correct answer:
A works twice as fast as B. If both of them can together finish a peice of work in 12 hours, then B alone can do it in
(a) 24 hours
(b) 27 hours
(c) 36 hours
(d) 18 hours

Answer 10:

(c) 36 hours

$Suppose B take s x hour s to complete the work . ∴ B' s 1 hour work = \frac{1}{x} A works twice as fast as B . ∴ A's 1 hour work = \frac{2}{x} \frac{1}{12} = \frac{1}{x} + \frac{2}{x} = \frac{3}{x} \Rightarrow x = 36 hours$

Q11 | Test Paper 13 | Class 8 | RS AGGARWAL | chapter 13 | Time and Work

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Question 11:

Fill in the blanks.
(i) A tap can fill a tank in 6 hours. The part of the tank filled in 1 hour is .........
(ii) A and B working together can finish a piece of work in 6 hours while A alone can do it in 9 hours. B alone can do it in ......... hours.
(iii) A can do a work in 16 hours and B alone can do it in 24 hours. If A, B and C working together can finish it in 8 hours, then C alone can finish it in ......... hours.
(iv) If A's one day's work is $\frac{3}{20}$ , then A can finish the whole work in ......... days.

Answer 11:

(i) A tap can fill a tank in 6 hours. In 1 hour, $\frac{1}{6}$ of the tank is filled.

(ii) 18 hours
$(A + B)' s 1 hour work = \frac{1}{6} A' s 1 hour work = \frac{1}{9} B' s 1 hour work = \frac{1}{6} - \frac{1}{9} = \frac{3 - 2}{18} = \frac{1}{18} Thus, B takes 18 hours to finish the work .$

(iii) 48 hours
$A' s 1 hour work = \frac{1}{16} B' s 1 hour work = \frac{1}{24} C' s 1 hour work = \frac{1}{x} (A + B + C)' s 1 hour work = \frac{1}{8} Therefore, \frac{1}{x} = \frac{1}{8} - \frac{1}{16} - \frac{1}{24} = \frac{6 - 3 - 2}{48} = \frac{1}{48} or, x = 48 hours Thus, C alone takes 48 hours to complete the wor k .$

(iv) The time for completion is the reciprocal of the work done in one day. Therefore, A can complete the whole work in $\frac{20}{3} = 6 \frac{2}{3} days$

RS Aggarwal solution class 8 chapter 13 Time and Work Exercise 13B

Exercise 13B

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Q1 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 1:

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A alone can do a piece of work in 10 days and B alone can do it in 15 days. In how many days will A and B together do the same work?
(a) 5 days
(b) 6 days
(c) 8 days
(d) 9 days

Answer 1:

(b) 6 days

$A can do a work in 10 days . A' s 1 day work = \frac{1}{10} B can do a work in 15 days . B' s 1 day work = \frac{1}{15} (A + B)' s 1 day work = \frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6} A and B together will take 6 days to complete the work .$

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Question 2:

Tick (✓) the correct answer:
A man can do a piece of work in 5 days. He and his son working together can finish it in 3 days. In how many days can the son do it alone?
(a) $6 \frac{1}{2}$ days
(b) 7 days
(c) $7 \frac{1}{2}$ days
(d) 8 days

Answer 2:

(c) $7 \frac{1}{2} d a y s$

$A man can do a work in 5 days . The man' s 1 day work = \frac{1}{5} The man and the son can do the work in 3 days . The man and his son' s 1 day work = \frac{1}{3} Let the son' s 1 day work be \frac{1}{x} . Therefore, \frac{1}{3} = \frac{1}{5} + \frac{1}{x} or, \frac{1}{x} = \frac{1}{3} - \frac{1}{5} = \frac{5 - 3}{15} = \frac{2}{15} x = \frac{15}{2} = 7 \frac{1}{2} days$

Q3 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 3:

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A can do a job in 16 days and B can do the same job in 12 days. With the help of C, they can finish the job in 6 days only. Then, C alone can finish it in
(a) 34 days
(b) 22 days
(c) 36 dyas
(d) 48 days

Answer 3:

(d) 48 days

$A can do a job in 16 days . B can do the job in 12 days . Suppose C can do the job in x days . A' s 1 day work = \frac{1}{16} B' s 1 day work = \frac{1}{12} C' s 1 day work = \frac{1}{x} A, B and C together can complete the work in 6 days . (A + B + C)' s 1 day work = \frac{1}{6} Therefore, \frac{1}{6} = \frac{1}{16} + \frac{1}{12} + \frac{1}{x} \Rightarrow \frac{1}{x} = \frac{1}{6} - \frac{1}{16} - \frac{1}{12} = \frac{8 - 3 - 4}{48} = \frac{1}{48} x = 48 Therefore, C alone can complete the job in 48 days .$

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Question 4:

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To complete a work, A takes 50% more time than B. If together they take 18 days to complete the work, how much time shall B take to do it?
(a) 30 days
(b) 35 days
(c) 40 days
(d) 45 days

Answer 4:

(a) 30 days

$L e t B t a k e x d a y s t o c o m p l e t e t h e w o r k . T h e n A t a k e s (x + \frac{50}{100} x) = 1.5 x A' s 1 d a y' s w o r k = \frac{1}{1.5 x} = \frac{2}{3 x} B' s 1 d a y' s w o r k = \frac{1}{x} (A + B) t a k e s 18 d a y s t o c o m p l e t e t h e w o r k . (A + B)' s 1 d a y' s n e t w o r k = \frac{1}{18} o r \frac{1}{18} = \frac{2}{3 x} + \frac{1}{x} \Rightarrow \frac{1}{18} = \frac{5}{3 x} B y c r o s s - m u l t i p l i c a t i o n, w e g e t : x = 30 d a y s ∴ B a l o n e w i l l t a k e 30 d a y s t o c o m p l e t e t h e w o r k .$

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Question 5:

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A works twice as fast as B. If both of them can together finish a piece of work in 12 days, then B alone can do it in
(a) 24 days
(b) 27 days
(c) 36 days
(d) 48 days

Answer 5:

$Then B takes 2 x days to complete the work .$

$A' s 1 day' s work = \frac{1}{x} B' s 1 day' s work = \frac{1}{2 x} A and B take 12 days to complete the work .$

$Net work done by (A + B) in 1 day = \frac{1}{12} = \frac{1}{x} + \frac{1}{2 x} = \frac{3}{2 x} \Rightarrow 2 x = 36 \Rightarrow x = 18 A can complete the work by himself in 18 days .$

$B will take 36 days, i . e ., twice as long as the time taken by A .$

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Question 6:

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A alone can finish a piece of work in 10 days which B alone can do in 15 days. If they work together and finish it, then out of total wages of Rs 3000, A will get
(a) Rs 1200
(b) Rs 1500
(c) Rs 1800
(d) Rs 2000

Answer 6:

(c) Rs. 1800

Since the wage distribution will follow the work distribution ratio, we have:

Work done by A in 1 day $= \frac{1}{10}$
Work done by B in 1 day $= \frac{1}{15}$

Net work done by (A+B) in 1 day $= \frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6}$

i.e., (A+B) will take 6 days to complete the work.

A's share of work in a day = $\frac{1}{10} \div \frac{1}{6} = \frac{1}{10} \times \frac{6}{1} = \frac{6}{10} = \frac{3}{5}$

∴ A's wage = $\frac{3}{5} \times 3000 = Rs 1800$

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Question 7:

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The rates of working of A and B are in the ratio 3 : 4. The number of days taken by them to finish the work are in the ratio
(a) 3 : 4
(b) 9 : 16
(c) 4 : 3
(d) 16 : 9

Answer 7:

(c) 4:3

The number of days taken for working is the reciprocal of the rate of work.

$i . e ., number of days taken = \frac{1}{rate of work} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

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Question 8:

Tick (✓) the correct answer:
A and B together can do a piece of work in 12 days; B and C can do it in 20 days while C and A can do it in 15 days. A, B and C all working together can do it in
(a) 6 days
(b) 9 days
(c) 10 days
(d) $10 \frac{1}{2}$ days

Answer 8:

(c) 10 days

$(A + B) can do a work in 12 days . (B + C) can do a work in 20 days . (C + A) can do a work in 15 days . Now, we have : Work done by (A + B) in 1 day = \frac{1}{12} Work done by (B + C) in 1 day = \frac{1}{20} Work done by (C + A) in 1 day = \frac{1}{15} Net work done by 2 (A + B + C) = \frac{1}{12} + \frac{1}{20} + \frac{1}{15} = \frac{5 + 3 + 4}{60} = \frac{12}{60} = \frac{1}{5} Net work done by (A + B + C) in 1 day = \frac{1}{10} ∴ If A, B and C work together, they will complete the work in 10 days .$

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Question 9:

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3 men or 5 women can do a work in 12 days. How long will 6 men and 5 women take to do it?
(a) 6 days
(b) 5 days
(c) 4 days
(d) 3 days

Answer 9:

(c) 4 days

$Three men can complete the work in 12 days . Thus, one man can complete the work in 36 days . Rate of work done by one man in 1 day = \frac{1}{36} Similarly, rate of work done by one woman in 1 day = \frac{1}{5 \times 12} = \frac{1}{60} Now, six men will do \frac{6}{36}, i . e ., \frac{1}{6} unit of work in a day . Five women will do \frac{5}{60}, i, e ., \frac{1}{12} unit of work in a day . ∴ Total work done in 1 day = \frac{1}{6} + \frac{1}{12} = \frac{1}{4} unit$

$Thus, six men and five women will take 4 days to complete the work .$

The work can be completed in 4 days.

Q10 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 10:

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A can do a piece of work in 15 days. B is 50% more efficient than A. B can finish it in
(a) 10 days
(b) $7 \frac{1}{2}$ days
(c) 12 days
(d) $10 \frac{1}{2}$ days

Answer 10:

(a) 10 days

$Work done by A in 1 day = \frac{1}{15} B is 50 % more efficient than A . ∴ Work done by B in 1 day = \frac{150}{100} \times \frac{1}{15} = \frac{1}{10} Thus, B can complete the work in 10 days .$

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Q11 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 11:

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A does 20% less work than B. If A can finish a piece of work in $7 \frac{1}{2}$ hours, then B can finish it in
(a) 5 hours
(b) $5 \frac{1}{2}$ hours
(c) 6 hours
(d) $6 \frac{1}{2}$ hours

Answer 11:

(c) 6 hours

$Time taken by A to finish the piece of work = 7 \frac{1}{2} hours = \frac{15}{2} hours Work done by A in 1 hour = \frac{2}{15} Let B take x hours to finish the work . Work done by B in 1 hours = \frac{1}{x} A can work 20 % less than B, or A can do \frac{4}{5} of B' s work .$

$Now, \frac{(\frac{4}{5})}{1} = \frac{(\frac{2}{15})}{(\frac{1}{x})} \Rightarrow \frac{4}{5} = \frac{2 x}{15} \Rightarrow x = \frac{15 \times 4}{5 \times 2} = 6 hours$

Q12 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 12:

Tick (✓) the correct answer:
A can do a piece of work in 20 days which B alone can do in 12 days. B worked at it for 9 days. A can finish the remaining work in
(a) 3 days
(b) 5 days
(c) 7 days
(d) 11 days

Answer 12:

(b) 5 days

$A can complete the work in 20 days . Work done by A in 1 day = \frac{1}{20} B can complete the work in 12 days . Work done by B in 1 day = \frac{1}{12} In 9 days, B completes \frac{9}{12}, i . e ., \frac{3}{4} of the work and leaves 1 - \frac{3}{4}, i . e ., \frac{1}{4} of the work undone . ∴ Time taken by A = \frac{1}{4} \div \frac{1}{20} = \frac{1}{4} \times 20 = 5 days$

Q13 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 13:

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A can do a piece of work in 25 days, which B alone can do in 20 days. A started the work and was joined by B after 10 days. The work lasted for
(a) $12 \frac{1}{2}$ days
(b) 15 days
(c) $16 \frac{2}{3}$ days
(d) 14 days

Answer 13:

$Work done by A in 1 day = \frac{1}{25} B can do the same work in 20 days .$

$Work done by B in 1 day = \frac{1}{20} A alone completes \frac{10}{25}, i, e ., \frac{2}{5} of the work in 10 days .$

$Now, work remaining = 1 - \frac{2}{5} = \frac{3}{5}$

$Work done by (A + B) in 1 day = \frac{1}{25} + \frac{1}{20} = \frac{9}{100}$

∴ Time taken if they work together=35÷9100=35×1009=203=623 days

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Question 14:

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Two pipes can fill a tank in 20 minutes and 30 minutes respectively. If both the pipes are opened simultaneously, then the tank will be filled in
(a) 10 minutes
(b) 12 minutes
(c) 15 minutes
(d) 25 minutes

Answer 14:

(b) 12 minutes

$First pipe can fill a tank in 20 minutes .$

$Second pipe can fill the tank in 30 minutes .$

$Part of tank filled by the first pipe in one minute = \frac{1}{20}$

$P art of tank filled by the second pipe in one minute \frac{1}{30}$

$Part of tank filled by both pipes in one minute = \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12}$

$Thus, it takes 12 minutes to fill the tank using both the pipes.$

Q15 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 15:

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A tap can fill a cistern in 8 hours and another tap can empty the full cistern in 16 hours. If both the taps are open, the time taken to fill the cistern is
(a) $5 \frac{1}{3}$ hours
(b) 10 hours
(c) 16 hours
(d) 20 hours

Answer 15:

(c) 16 hours

$A tap can fill a cistern in 8 hours . Part of cistern filled in one hour = \frac{1}{8} A tap can empty the cistern in 16 hours . Part of cistern emptied in one hour = - \frac{1}{16} (negative sign shows that the cistern is being drained) ∴ Part of cistern filled in one hour = \frac{1}{8} - \frac{1}{16} = \frac{1}{16} Time required to fill the cistern = 16 hours$

Q16 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 16:

Tick (✓) the correct answer:
A pump can fill a tank in 2 hours. Due to a leak in the tank it takes $2 \frac{1}{3}$ hours to fill the tank. The leak can empty the full tank in
(a) $2 \frac{1}{3}$ hours
(b) 7 hours
(c) 8 hours
(d) 14 hours

Answer 16:

(d) 14 hours

$A pump can fill a tank in 2 hours . Part of the tank filled by the pump in one hour = \frac{1}{2} Suppose the leak empties a full tank in x hours . Part of the tank emptied by the leak in one hour = - \frac{1}{x} Part of tank filled in one hour = \frac{1}{2} - \frac{1}{x} = \frac{3}{7} (given) \frac{1}{x} = \frac{1}{2} - \frac{3}{7} = \frac{7 - 6}{14} = \frac{1}{14} x = 14 hours$

Q17 | Ex-13B | Class 8 | RS AGGARWAL | chapter 13 | Time and Work | myhelper

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Question 17:

Tick (✓) the correct answer:
Two pipes can fill a tank in 10 hours and 12 hours respectively, while a third pipe empties the full tank in 20 hours. If all the three pipes operate simultaneously, in how much time will the tank be full?
(a) 7 hrs 15 min
(b) 7 hrs 30 min
(c) 7 hrs 45 min
(d) 8 hrs

Answer 17:

(b) 7 hours 30 minutes

$Part of the tank filled by the first pipe in one hour = \frac{1}{10} Part of the tank filled by the second pipe in one hour = \frac{1}{12} Part of the tank filled by the third pipe in one hour = \frac{- 1}{20} Part of the tank filled by three pipes in one hour = \frac{1}{10} + \frac{1}{12} - \frac{1}{20} = \frac{2}{15} Total time taken to fill the tank = \frac{15}{2} hrs = 7 hours 30 minutes$